1. The Normal Models/Distributions

2. Normal Distribution Parameters The first parameter specified here is shape: symmetric, unimodal, bell shaped. (A formula defines the exact shape.) For a Normal variable the mean  and standard deviation  are needed to proceed. All else follows from knowing these three things.

3. Z-scores (Standardized Values) Z measures “how many standard deviations from the mean” whether above (+) or below (-) the mean This is the same formula, solved for X. You are required to know these formulas.

4. Example The gestation period (length of pregnancy) for women is Normally distributed with mean  = 270 days and standard deviation  = 11 days. Draw a picture!

5. Example Use the mean to get started.

6. Example On the templates, consecutive tick marks are one standard deviation apart.

7. Example

8. A reference to getting started with Normal distributions. About 68% of the data fall within 1 standard deviation of the mean:    68 – 95 – 99.7 Rule also known as the Empirical Rule

9. Example 0.6827

10. Empirical (68 – 95 – 99.7) Rule A reference to getting started with Normal distributions. About 95% (all but roughly 1 out of 20) of the data fall within 2 standard deviations of the mean:   2 

11. Example 0.9544

12. Empirical (68 – 95 – 99.7) Rule A reference to getting started with Normal distributions. About 99.7% (all but roughly 1 in 370) of the data fall within 3 standard deviations of the mean:   3 

13. Example 0.9973

14. Data more than 3  from  0.0027 or about 1 in 370 data values (on average) fall outside   3 . Technically the minimum and maximum are -  and +  (infinite). Values far to the right and left have very very very very small likelihood.

15. Beyond  + 3  0.00135 =1/741 values more than 3  above  0.0000317 =1/31574 values more than 4  above  If the curve were drawn 1 mm high at 6  from , then it would be drawn 65.66 km high at .

16. Example The Z scale “goes with” the data scale. Z = # of standard deviations from the mean.

17. Example Determine the probability a pregnancy lasts at least 274 days (that’s 9 months).

18. Finding Probabilities To solve a problem asking for a probability with for a Normal distribution: Step 1. Draw a picture identifying the relevant area.

19. Example Determine the probability a pregnancy lasts at least 274 days.

20. Example Determine the probability a pregnancy lasts at least 274 days.

21. Example Determine the probability a pregnancy lasts at least 274 days. Area = Probability

22. Finding Probabilities To solve a problem asking for a probability with for a Normal distribution: Step 2. Convert the data value to the Z scale. Data = 274  = 270  = 11 days 274 is 0.36 standard deviations above the mean

23. Example Determine the probability a pregnancy lasts at least 274 days.

24. Finding Probabilities To solve a problem asking for a probability with for a Normal distribution: Step 3. Use the Table (A-2 in the book) or computer or calculator to determine the percentile for the value. A Z-score of 0.36 matches with a probability of 0.6406. That is: Z = 0.36 is the 64.06 percentile.

25. Example Determine the probability a pregnancy lasts at least 274 days. 0.6406

26. Finding Probabilities To solve a problem asking for a probability with for a Normal distribution: Step 4. Problems asking for the probability of a result above or between will require further operations, because of the way percentiles are oriented.

27. Example 0.6406 1 – 0.6406 = 0.3594 coincidence

28. Example - Solution 0.3594 is the probability that a pregnancy lasts at least 274 days. That is: 35.94% of all pregnancies last at least 274 days. (Probability calculations always apply to populations.) 274 days is the 64 percentile of gestation times.

29. Example The gestation period (length of pregnancy) for women is Normally distributed with mean  = 270 days and standard deviation  = 11 days.

30. Example A women gave birth to a child on October 2, 2007. She identifies a man as the father. The case is in court. The man testifies he is not the father, and documents that he was out of the country from December 19, 2006 to January 30, 2007. December 19 is 287 days before October 2. January 30 is 245 days before October 2. What is the probability a pregnancy lasts between 245 and 287 days? (If it’s high, then we have evidence – in the data – supporting the man’s claim that he is not the father.)

31. Example What’s the probability the pregnancy lasted between 245 and 287 days?

32. Example What’s the probability the pregnancy lasted between 245 and 287 days?

33. Example What’s the probability the pregnancy lasted between 245 and 287 days? Area = Probability

34. Convert both data values to the Z scale. 245 is 2.27 standard deviations below the mean 287 is 1.55 standard deviations above the mean Data = 245Data = 287

35. Example What’s the probability the pregnancy lasted between 145 and 287 days? Area = Probability

36. Example What’s the probability the pregnancy lasted between 145 and 287 days? 0.9394 = Area below 287 (1.55)

37. Example What’s the probability the pregnancy lasted between 145 and 287 days? 0.9394 = Area below 287 (1.55) 0.0116 = Area below 245 (-2.27)

38. Example What’s the probability the pregnancy lasted between 145 and 287 days? 0.9394 = Area below 287 (1.55) 0.0116 = Area below 245 (-2.27) 0.9278 = Area between 245 and 287

39. Example - Solution 0.9278 is the probability that a pregnancy lasts between 245 and 287 days. We found 245 to be the 1.16 percentile of pregnancy lengths. We found 287 to be the 93.94 percentile.

40. Example - Solution 0.9266 is the probability that the woman’s pregnancy had a length that proves the man’s innocence. Note: This is not the probability the man is not the father. In our way of thinking about things, there is no such thing as “the probability the man is the father.” Either he is or is not the father. (We just don’t know.)

41. Example Determine the first quartile of the pregnancy lengths. First draw a picture.

42. Example Determine the first quartile. 0.2500 = Area

43. Example Determine the first quartile of the pregnancy lengths. Notice that the Z value will be negative.

44. Example Determine the first quartile. 0.2500

45. Example Determine the first quartile of the pregnancy lengths. In problems like these – where the probability is given – the steps of the previous problems are reversed.

46. Example The steps of the previous problems are reversed. From the given probability / percentile, determine the Z-score. Since 0.2500 is the probability of a result less than Q1: Z  -0.67 (Better tools give -0.674.)

47. Example Determine the first quartile. 0.2500

48. Example You have the Z score (–0.67). Use either of the given equations to obtain the data value X.

49. Example Determine the first quartile. 0.2500 Q 1 = 262.63

50. Example You have the Z score (–0.67). Use either of the given equations to obtain the data value X.

51. Example The first quartile is 270 – 7.37 = 262.63. (7.37 below the mean) The median is 270. The third quartile is 270 + 7.37 = 277.37. (7.37 above the mean) The IQR is 2(7.37) = 14.74. SD / IQR = 11 / 14.74 = 0.746 For any Normal distribution, the SD is about 3/4ths the IQR.

52. Example At one hospital, women whose pregnancy falls into the top 2% of lengths are recommended for inducing. At what time (in days) is a woman recommended for inducing? In other words… Find the length that separates the longest 2% from the shortest 98% of pregnancies.

53. Example Find the length that separates the longest 2% from the shortest 98% of pregnancies.

54. Example Find the length that separates the longest 2% from the shortest 98% of pregnancies. Area = 0.98 Area = 0.02

55. Example Find the length that separates the longest 2% from the shortest 98% of pregnancies. Area = 0.98 Area = 0.02 X =  + Z   X = 270 + 2.054  11

56. Example Find the length that separates the longest 2% from the shortest 98% of pregnancies. Area = 0.98 Area = 0.02 X =  + Z   X = 292.594

57. Example At one hospital, women whose pregnancy falls into the top 2% of lengths are recommended for inducing. At what time (in days) is a woman recommended for inducing? ANS: A woman should be induced at 292.6 days. (This is the 98 th percentile of the distribution.)

58. Percentiles / Z scores Example 25 th percentileZ = -0.67  X = 262.6 Example 98 th percentileZ = 2.05  X = 292.6

59. Percentiles / Z scores Example 25 th percentileZ = -0.67  X = 262.6 In general for any variable with Normal distribution The 25 th percentile is 0.67 standard deviations below the mean. Example 98 th percentileZ = 2.05  X = 292.6 In general for any variable with Normal distribution The 98 th percentile is 2.05 standard deviations above the mean.

60. IQR Rule of Thumb For a Normal distribution,   ¾ IQR. The standard deviation is fairly close to 75% of the interquartile range. 0.6745  2/3 IQR(Z)  (4/3) IQR  (4/3) SD SD  (3/4) IQR